The circumference of a circle is 10% more than the perimeter of a square. If the difference between the area of the circle and that of the square is 216 cm2, how much does the diagonal of the square measure ?
(A) $$14\sqrt 2 $$
(B) $$14$$
(C) $$20$$
(D) $$20\sqrt 2 $$
Solution:
Let the radius of circle be r cm and side of square be a cm Then circumference of circle = $$2\pi r$$ and Perimeter of square = 4a According to the question, $$\eqalign{ & 2\pi r = 4a \times \frac{{110}}{{100}} \cr & \Rightarrow 2\pi r = \frac{{44a}}{{10}} \cr & \Rightarrow r = \frac{{44a}}{{2\pi \times 10}} \cr & \Rightarrow r = \frac{{11a}}{{5\pi }} \cr & \Rightarrow a = \frac{{5\pi r}}{{11 }} . . . . .(i) \cr & {\text{Also, }}\pi {r^2} - {a^2} \Rightarrow 216 \cr} $$ $$ \Rightarrow \pi {r^2} - \frac{{25\pi {r^2}}}{{121}} = 216$$ $$\left[ {{\text{from equation (i)}}} \right]$$ $$\eqalign{ & \Rightarrow \frac{{121\pi {r^2} - 25\pi {r^2}}}{{121}} = 216 \cr & \Rightarrow {r^2}\left[ {121\pi - 25{\pi ^2}} \right] = 26136 \cr} $$ $$ \Rightarrow {r^2}\left[ {121 \times \frac{{22}}{7} - 25 \times \frac{{22}}{7} \times \frac{{22}}{7}} \right]$$ $$ = 26136$$ $$\eqalign{ & \Rightarrow {r^2}\left[ {\frac{{2662}}{7} - \frac{{12100}}{{49}}} \right] = 26136 \cr & \Rightarrow {r^2}\left[ {\frac{{6534}}{{49}}} \right] = 26136 \cr & \Rightarrow {r^2} = \frac{{26136 \times 49}}{{6534}} \cr & \Rightarrow {r^2} = 196 \cr & \Rightarrow r = 14\,cm \cr & \therefore a = \frac{{5\pi r}}{{11}} = 5 \times \frac{{22}}{7} \times \frac{{14}}{{11}} = 20\,cm \cr} $$ Hence, diagonal of square $$ = \sqrt 2 a = 20\sqrt 2 \,cm$$
72.
An equilateral triangle is described on the diagonal of a square. What is the ratio of the area of the triangle to that of the square ?
(A) 2 : $$\sqrt 3 $$
(B) 4 : $$\sqrt 3 $$
(C) $$\sqrt 3 $$ : 2
(D) $$\sqrt 3 $$ : 4
Solution:
Let the side of the square be a cm Then, the length of its diagonal = $$\sqrt 2 $$ a cm Area of equilateral triangle with side : $$\eqalign{ & = \sqrt 2 a \cr & = \frac{{\sqrt 3 }}{4} \times {\left( {\sqrt 2 a} \right)^2} \cr & = \frac{{\sqrt 3 {a^2}}}{2} \cr} $$ ∴ Required ratio : $$\eqalign{ & = \frac{{\sqrt 3 {a^2}}}{2}:{a^2} \cr & = \sqrt 3 :2 \cr} $$
73.
A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:
Of the two square fields, the area of one is 1 hectare while the other one is broader by 1%. The difference in their areas is :
(A) 100 m2
(B) 101 m2
(C) 200 m2
(D) 201 m2
Solution:
Area = 1 hect. = 10000 sq. m Side = $$\sqrt {10000} $$ m = 100 m Side of the other square = 101 m Difference in their areas : = [(101)2 - (100)2] m2 = [(101 + 100) (101 - 100)] m2 = 201 m2
75.
A plate on square base made of brass is of length x cm and width 1 mm. The plate weights 4725 gm. If 1 cubic cm cm of brass weight 8.4 grams, then the value of x is :
(A) 75
(B) 76
(C) 72
(D) 74
Solution:
Given length and width of a square base plate of brass is x cm and 1 mm Volume of the plate of square base = Area of base × height $$\eqalign{ & = {x^2} \times \frac{1}{{10}} \cr & = \frac{{{x^2}}}{{10}}\,cu.cm. \cr} $$ According to the question,$$\eqalign{ & \Rightarrow \frac{{{x^2}}}{{10}} \times 8.4 = 4725 \cr & \Rightarrow {x^2} = \frac{{4725 \times 10}}{{8.4}} \cr & \Rightarrow {x^2} = 5625 \cr & \Rightarrow x = \sqrt {5625} = 75\,cm \cr} $$
76.
The area of a rectangle is 252 cm2 and its length and breadth are in the ratio of 9 : 7 respectively. What is its perimeter ?
(A) 64 cm
(B) 68 cm
(C) 96 cm
(D) 128 cm
Solution:
Let the length and breadth of the rectangle be (9x) cm and (7x) cm respectively. Then, $$\eqalign{ & 9x \times 7x = 252 \cr & \Rightarrow 63{x^2} = 252 \cr & \Rightarrow {x^2} = 4 \cr & \Rightarrow x = 2 \cr} $$ So, length = 18 cm, breadth = 14 cm ∴ Perimeter : = 2(18 + 14) cm = 64 cm
77.
The length of a rectangular blackboard is 8 m more than its breadth. If its length is increased by 7 m and its breadth is decreased by 4 m, its area remains unchanged. The length and breadth of the rectangular blackboard is :
(A) 24 m, 16 m
(B) 20 m, 24 m
(C) 28 m, 16 m
(D) 28 m, 20 m
Solution:
Let the breadth = x cm Then, length = (x + 8) m $$\eqalign{ & \therefore \left( {x + 8} \right)x = \left( {x + 15} \right)\left( {x - 4} \right) \cr & \Rightarrow {x^2} + 8x = {x^2} + 11x - 60 \cr & \Rightarrow x = 20 \cr} $$ So, length = 28 m and breadth = 20 m
78.
The circumference of a circular ground is 88 metres. A strip of land, 3 metres wide, inside and along the circumference of the ground is to be levelled. What is the budgeted expenditure if the levelling costs Rs. 7 per square metre ?
(A) Rs. 1050
(B) Rs. 1125
(C) Rs. 1325
(D) Rs. 1650
Solution:
Let the radius of the ground be R metres. Then, $$\eqalign{ & \Rightarrow 2\pi R = 88 \cr & \Rightarrow R = \left( {\frac{{88 \times 7}}{{2 \times 22}}} \right) \cr & \Rightarrow R = 14\,m \cr} $$ Area of land strip : $$\eqalign{ & = \pi \left[ {{{\left( {14} \right)}^2} - {{\left( {11} \right)}^2}} \right]{m^2} \cr & = \left( {\frac{{22}}{7} \times 25 \times 3} \right){m^2} \cr & = \left( {\frac{{1650}}{7}} \right){m^2} \cr} $$ ∴ Cost of levelling : $$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {\frac{{1650}}{7} \times 7} \right) \cr & = {\text{Rs}}{\text{. 1650}} \cr} $$
79.
A small disc of radius r is cut out from a disc of radius R. The weight of the disc which now has a hole in it, is reduced to $$\frac{{24}}{{25}}$$ of the original weight. If R = xr, what is the value of x ?
(A) 4
(B) 4.5
(C) 24
(D) 25
Solution:
Answer & Solution Answer: Option E Solution: Since weight of the disc is proportional to its area, we have : $$\eqalign{ & \pi \left( {{R^2} - {r^2}} \right) = \frac{{24}}{{25}}\pi {R^2} \cr & \Rightarrow {R^2} - {r^2} = \frac{{24}}{{25}}{R^2} \cr & \Rightarrow {r^2} = \frac{1}{{25}}{R^2} \cr & \Rightarrow {R^2} = 25{r^2} \cr & \Rightarrow R = 5r \cr} $$
80.
If the diagonal and the area of a rectangle are 25 m2 and 168 m2, what is the length of the rectangle ?